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Alex73 [517]
3 years ago
8

What are the factor that affect the efficiency of a pulley​

Physics
1 answer:
alukav5142 [94]3 years ago
8 0

Answer:

Tension in the chains - In a chain drive, technically, you have a closed-chain (which has no end) going around 2 pulley or gears; looking closely you have 2 parallel chains going in opposite direction. If kept in horizontal direction, the one below the other is the slack side and the other the tight side. The tension on the upper or tight side is more than the slack side. So you need to keep in mind to keep your chain drive tight so that there is no loss or rotation or lags.

Sizes of the pulley/gear - The chain will be warped around a pair of pulley or gear. The sizes of these pulley/gear will also determine the efficiency of the chain drive (consider one big and one small)

Number of pulley/gear - If the number of pulley/gear is more and chain wrapped on it with little complexity will result in decrease in efficiency because of extra tension.

Length of the chain drive - You cannot have much too long chain drive. It will make your slack side more heavy because the end are further away. You have to apply more power and possibilities of lag increases decreasing efficiency. In an ideal situation, this won't happen, but this world isn't ideal.

Friction between chains & pulley/gear - If you have studied gears (involving its teeth), you will come to know that there is friction offered on the two meeting surfaces.

Angle of contact - This would have been explained better with a diagram. Although, if you are familiar with the terms you won't have difficulty understanding. Angle of contact is the angle the chain forms with the pulley/gear at the point of contact with the center of the pulley. The angle of contact should not be too small, or else the things will be slippery.

Explanation:

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Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

           y is the difference in weight which is   =1.10-0.50=0.6m

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

                       = 8.17 N/m

The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

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